If (x – 5) is the HCF of x² – x – p and x² – qx – 10, then what is the value of (p + q)?

Since (x – 5) is a factor, substituting x = 5 in both polynomials gives 0. For x² – x – p: 25 – 5 – p = 0, so p = 20. For x² – qx – 10: 25 – 5q – 10 = 0, so 5q = 15, q = 3. Therefore p + q = 20 + 3 = 23.